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3 years ago

What is a reasonable measurement for the distance to Neptune?

Physics

2 answers:

IRINA_888 [86]3 years ago

5 0

Answer:

30 kilometers is a reasonable measurement

Triss [41]3 years ago

4 0

Answer:

30 kilometers , that's the answer

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A 1,160 kg satellite orbits earth with a tangential speed of 7,446 m/s. if the satellite experiences a centripetal force of 8,95

Mkey [24]

The satellite is 8.02 × 10⁵ m above Earth's surface.

Let H be the height above the surface of the Earth; since we know that the satellite is rotating around the Earth due to the gravitational pull of the planet, we may assert

Procedure to solve:

F = mv²/R+H

H = mv²/F - R

H = (1160 × 7446²/8955 - 6.38 × 10⁶)

M = 8.02 × 10⁵ m

About centripetal force:

The force applied to an item that is in velocity of curved motion that is pointed toward the axis of rotation or the centre of curvature is known as a centripetal force.

The centripetal force formula is given as the product of mass (in kg) and tangential velocity (in meters per second) squared, divided by the radius (in meters) that implies that on doubling the tangential velocity, the centripetal force will be quadrupled. Mathematically it is written as:

F = mv²/r

Learn more about velocity here:

brainly.com/question/18084516

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6 0

2 years ago

A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w

steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

$h = L - L\cos{31°} = L(1 - \cos{31°})$

so its initial gravitational potential energy $U_0$ is

$U = mgh = mgL(1 - \cos{31°})$

$\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})$

$\:\:\:\:\:=0.23\:\text{J}$

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

$\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U$

We know that the initial kinetic energy $K_0,$ as well as its final gravitational potential energy $U$ are zero so we can write the conservation law as

$mgL(1 - \cos{31°}) = \frac{1}{2}mv^2$

Note that the mass gets cancelled out and then we solve for the velocity v as

$v = \sqrt{2gL(1 - \cos{31°})}$

$\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}$

$\:\:\:\:\:= 1.3\:\text{m/s}$

5 0

3 years ago

Read 2 more answers

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst

Archy [21]

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

$E = \frac{1}{2}mv^{2}$

So, $v = \sqrt{\frac{2E}{m}}$ .... (1)

Now,

$r = \frac{mv}{Bq}$

Substitute the value of v from equation (1), we get

$r = \frac{\sqrt{2mE}}{Bq}$

Let the radius of the alpha particle is r2.

For proton

So, $r_{1} = \frac{\sqrt{2m_{1}E}}{Bq_{1}}$ ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_{2} = \frac{\sqrt{2m_{2}E}}{Bq_{2}}$ ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$\frac{r_{1}}{r_{2}}=\frac{q_{2}}{q_{1}}\times \sqrt{\frac{m_{1}}{m_{2}}}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$\frac{r_{1}}{r_{2}}=\frac{2q}}{q}}\times \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

8 0

3 years ago

Read 2 more answers

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

Which conditions are necessary for clouds to form?

VikaD [51]

Answer: Some of the thing that are important for clouds to form are: Moisture - There must be sufficient water vapor in the air for a cloud to form. Cooling air - The air temperature must decrease enough for water vapor to condense

Explanation:

7 0

3 years ago

Read 2 more answers